1. Chapter 16 Lesson 2 Solubility and Complex Ion Equilibria

2. Chapter 16 Table of Contents Copyright © Cengage Learning. All rights reserved 2 16.1Solubility Equilibria and the Solubility Product 16.2 Precipitation and Qualitative Analysis 16.3 Equilibria Involving Complex Ions

3. 3 Precipitation Precipitation is merely another way of looking at solubility equilibrium. –Rather than considering how much of a substance will dissolve, we ask: Will precipitation occur for a given starting ion concentration?

4. 4 Criteria for Precipitation To determine whether an equilibrium system will go in the forward or reverse direction requires that we evaluate the reaction quotient, Q (or IP). –To predict the direction of reaction, you compare Q with K sp. –The reaction quotient has the same form as the K sp expression, but the concentrations of products are starting values not necessarily saturated conc.

5. 5 Criteria for Precipitation –Consider the following equilibrium. H2OH2O where initial concentration is denoted by i.

6. 6 Criteria for Precipitation –If Q = K sp, the solution is just saturated with ions and any additional solid will not dissolve in solution but instead will precipitate out. –If Q < K sp, the solution is unsaturated and more solid can be dissolved in the solution; no precipitate forms. Shift right increase Q –If Q > K sp,, the solution is supersaturated meaning the solution contains a higher concentration of ions than possible at equilibrium; a precipitate will form. Shift to left decrease Q solid ions Q = [ions]

7. 7 The concentration of calcium ion in blood plasma is 0.0025 M. If the concentration of oxalate ion is 1.0 x 10 -5 M, do you expect calcium oxalate to precipitate? K sp for calcium oxalate is 2.3 x 10 -9. –The ion product quotient, Q c, is: –This value is larger than the K sp, so you expect precipitation to occur past saturation point. 0.0025 M 1.0 x 10 -5 M

8. 8 The concentration of lead ion is 0.25 M. If the concentration of chloride ion is 0.0060 M, do you expect lead chloride to precipitate? K sp for lead chloride is 1.7 x 10 -5. PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) Note: did not double chloride concentration Q < K sp = 1.7 x 10 -5, indicating that a no precipitate will form (below the saturation point). 0.25 M 0.0060 M

9. 9 example: A student mixes 0.200 L of 0.0060 M Sr(NO 3 ) 2 solution with 0.100 L of 0.015 M K 2 CrO 4 solution to give a final volume of 0.300L. Will a precipitate form under these conditions? K sp SrCrO 4 = 3.6 x 10 -5 Sr(NO 3 ) 2 (aq) + K 2 CrO 4 (aq)  SrCrO 4 (s) + 2 KNO 3 (aq) SrCrO 4 (s) Sr 2+ (aq) + CrO 4 2- (aq) we find that Q (2.0 x 10 -5 ) < K sp (3.6 x 10 -5 ) indicating that no precipitate will form (below saturation point) 0.0040 M 0.0050 M (aq)

10. 10 Selective Precipitation Selective precipitation is the technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth. –For example, when you slowly add potassium chromate, K 2 CrO 4, to a solution containing Ba 2+ and Sr 2+, barium chromate precipitates first due to its lower solubility than SrCrO 4.

11. 11 –After most of the Ba 2+ ion has precipitated, strontium chromate begins to precipitate. Selective Precipitation –It is therefore possible to separate Ba 2+ from Sr 2+ by selective precipitation using K 2 CrO 4. –Take advantage in qual/quan type analysis.

12. 12 Effect of pH on Solubility Sometimes it is necessary to account for other reactions that aqueous ions might undergo. –For example, if the anion is the conjugate base of a weak acid, it will react with H 3 O +. –You should expect the solubility to be affected by pH. By adding and complexing out ions you can affect the pH of solution which could affect ppt reactions.

13. 13 Effect of pH on Solubility –Consider the following equilibrium. H2OH2O –Because the oxalate ion is conjugate base, it will react with H 3 O + (added acid to lower pH). H2OH2O –According to Le Chatelier’s principle, as C 2 O 4 2- ion is removed by the reaction with H 3 O +, more calcium oxalate dissolves (increase solubility). –Therefore, you expect calcium oxalate to be more soluble in acidic solution (lower pH) than in pure water. The acidity will react with the oxalate and shift the equil toward the right and allow more calcium oxalate to dissolve. s by pH

14. 14 Separation of Metal Ions by Sulfide Precipitation Many metal sulfides are insoluble in water but dissolve in acidic solution. –Qualitative analysis uses this change in solubility of the metal sulfides with pH to separate a mixture of metal ions. –By adjusting the pH in an aqueous solution of H 2 S, you adjust the sulfide concentration to precipitate the least soluble metal sulfide first. – Typically do some qual experiment similar in lab.

15. 15 Qualitative Analysis Qualitative analysis involves the determination of the identity of substances present in a mixture. –In the qualitative analysis scheme for metal ions, a cation is usually detected by the presence of a characteristic precipitate. –Next slide shows a figure that summarizes how metal ions in an aqueous solution are separated into five analytical groups.

16. 16 Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005.

17. Section 16.2 Atomic MassesPrecipitation and Qualitative Analysis Return to TOC Copyright © Cengage Learning. All rights reserved 17 Separating the Common Cations by Selective Precipitation

18. Section 16.2 Atomic MassesPrecipitation and Qualitative Analysis Return to TOC Copyright © Cengage Learning. All rights reserved 18 Separation of Cu 2+ and Hg 2+ from Ni 2+ and Mn 2+ using H 2 S At a low pH, [S 2– ] is relatively low and only the very insoluble HgS and CuS precipitate. When OH – is added to lower [H + ], the value of [S 2– ] increases, and MnS and NiS precipitate.

19. Section 16.2 Atomic MassesPrecipitation and Qualitative Analysis Return to TOC Copyright © Cengage Learning. All rights reserved 19 Separation of Cu 2+ and Hg 2+ from Ni 2+ and Mn 2+ using H 2 S

20. Section 16.3 The MoleEquilibria Involving Complex Ions Return to TOC Copyright © Cengage Learning. All rights reserved 20 Complex Ion Equilibria Charged species consisting of a metal ion surrounded by ligands.  Ligand: Lewis base Formation (stability) constant.  Equilibrium constant for each step of the formation of a complex ion by the addition of an individual ligand to a metal ion or complex ion in aqueous solution.

21. Section 16.3 The MoleEquilibria Involving Complex Ions Return to TOC Copyright © Cengage Learning. All rights reserved 21 Complex Ion Equilibria Be 2+ (aq) + F – (aq) BeF + (aq) K 1 = 7.9 x 10 4 BeF + (aq) + F – (aq) BeF 2 (aq) K 2 = 5.8 x 10 3 BeF 2 (aq) + F – (aq) BeF 3 – (aq) K 3 = 6.1 x 10 2 BeF 3 – (aq) + F – (aq) BeF 4 2– (aq) K 4 = 2.7 x 10 1

22. Section 16.3 The MoleEquilibria Involving Complex Ions Return to TOC Copyright © Cengage Learning. All rights reserved 22 Complex Ions and Solubility Two strategies for dissolving a water–insoluble ionic solid.  If the anion of the solid is a good base, the solubility is greatly increased by acidifying the solution.  In cases where the anion is not sufficiently basic, the ionic solid often can be dissolved in a solution containing a ligand that forms stable complex ions with its cation.

23. Section 16.3 The MoleEquilibria Involving Complex Ions Return to TOC How can one dissolve silver chloride and pull the rxn to the right even though chloride is a weak base? AgCl (s) Ag + (aq) + Cl - (aq) Copyright © Cengage Learning. All rights reserved 23

24. Section 16.3 The MoleEquilibria Involving Complex Ions Return to TOC Copyright © Cengage Learning. All rights reserved 24 Concept Check Calculate the solubility of silver chloride in 10.0 M ammonia given the following information: K sp (AgCl) = 1.6 x 10 – 10 Ag + + NH 3 AgNH 3 + K = 2.1 x 10 3 AgNH 3 + + NH 3 Ag(NH 3 ) 2 + K = 8.2 x 10 3 0.48 M Calculate the concentration of NH 3 in the final equilibrium mixture. 9.0 M

25. Formation Constants for Complex Ions The solution of a slightly soluble salt increase when one of its ions can be changed to a soluble complex ion. AgBr (s) Ag + + Br - K sp = 5.0 x 10 -13 Add NH 3 Ag + + 2NH 3 Ag(NH 3 ) 2 + kf orm = 1.6 x 10 7

26. Formation Constants for Complex Ions The very soluble silver complex ion removes Ag+ from the solution and shifts the equilibrium to the right increasing the solubility of AgCl. AgBr + 2NH 3 Ag(NH 3 ) 2+ + Br - K c = 8.0 x 10 -6 = [Ag(NH 3 ) 2+ ][Br - ] [NH 3 ] 2 K c = k form x k sp = (1.6 x 10 7 )(5.0x10 -13 ) = 8.0 x 10 -6

27. Example How many moles of AgBr can dissolve in 1 L of 1.0 M NH 3 ? AgBr + 2NH 3 Ag(NH 3 ) 2+ + Br – 1.000 -2x+x+x 1.0-2xxx Kc = x 2 (1.0-2x) 2 = 8.0x10 -6 X = 2.8x10 -3, 2.8 x 10 -3 mol of AgBr dissolves in 1L of NH 3

28. 28 Relative Solubility Comparing solubilities: Which is most soluble in water? CaCO 3 K sp = 3.8 x 10 -9 K sp = [Ca 2+ ] [CO 3 2- ] =s 2 = 3.8 x 10 -9 s = 6.2 x 10 -5 M AgBr K sp = 5.0 x 10 -13 K sp = [Ag + ] [Br - ] = s 2 = 5.0 x 10 -13 s = 7.1 x 10 -7 M CaF 2 K sp = 3.4 x 10 -11 K sp = [Ca 2+ ] [F - ] 2 =(s)(2s) 2 = 4s 3 = 3.4 x 10 -11 s = 2.0 x 10 -4 M CaCO 3 (s) Ca 2+ (aq) + CO 3 2- (aq) s AgBr (s) Ag + (aq) Br - (aq) s CaF 2 (s) Ca 2+ (aq) + 2 F - (aq) s 2 s